0
$\begingroup$

Here is the statement of the wikipedia page for mathematical symbols and i'm struggling to understand it in context;

Statement: Every finite, non-empty, ordered set has a largest element. Otherwise, let's assume that X is a finite, non-empty, ordered set with no largest element. Then, for some $x_1$ is a member of $X$, there exists an $x_2$ in $X$ with $x_1 < x_2$, but then there's also an $x_3$ in $X$ with $x_2 < x_3$, and so on. Thus, $x_1, x_2, x_3, ...$ are distinct elements in X. ↯ X is finite.

My thinking:

Every finite, non-empty, order set has a largest element. (True, a non-empty set has a element which has a quantity.)

Otherwise X is finite, non-empty, ordered with no largest element. (This can't happen, so it's unnecessary to make this statement?)

Then, for some $x_1$ in $X$ there exists an $x_2$ in $X$ with $x_1 < x_2$ (Well maybe not? if X is of length 1; Is this the contradictory statement?)

It then goes on to say this holds true for all $x_{n+1}$ (If what was stated before then this can't hold?)

My question is what does this ↯ mean in this statement, at what point is the contradiction? Isn't that kind of ambiguous in a mathematical proof?

Can someone shed some light, i'm getting all kinds of confused.

$\endgroup$
1
$\begingroup$

You should read this as "Contradiction, because $X$ is finite."

The assumption is that X is finite and has no largest element, and argument proceeds to show why that doesn't work. (Since it's finite, the chain $x_1 < x_2 < x_3 < ...$ must stop at some $x_n$)

$\endgroup$
  • $\begingroup$ That makes so much more sense.... Thanks... Feel like a bit of an idiot haha $\endgroup$ – Alex White Dec 16 '17 at 10:02
  • $\begingroup$ @AlexWhite Always ask questions. :) $\endgroup$ – naslundx Dec 16 '17 at 10:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.